dream_weaver Posted March 5, 2013 Report Share Posted March 5, 2013 Fermat's Last Theorem Unlocking the Secret of an Ancient Mathematical Problem. Amir D. Aczel Your knowledge of math won't necessarily be challenged by this short 137 page hardcover edition. It presents a look at the mathematicians who provided various keys to Professor Andrew Wiles 1993 confirmation of a note made in a margin three centuries earlier. Pierre de Fermat had written around 1637 inthe "Arithmetica" written by the Greek mathematician Diophantus, next to a problem on breaking down a squared number into two squares: On the other hand, it is impossible to separate a cube into two cubes, or a biquadrate into two biquadrats, or generally any power except a square into two powers with the same exponent. I have discovered a truly marvelous proof of thes, which, however, the margin is not large enough to contain. Fermat broke ground developing the main ideas of calculus 13 years before Newton was born. Amir suggests that the origins of A^{2}+B^{2}=C^{2} being resolved for integers go back to at least Circa 2000 B.C. with connections to wealth, land surveying, and religion. He spends about 1/3 of the book various observations and methods that were developed and available to Fermet. The book is filled with thumbnail biographies of key mathematicians where available, and what their contributions were - speckled with a few mathematical representations, all taking place after Fermet, and viewed as instrumental for Andrew Wiles to have provided the presentation he did. Amir closes the book withthe statement: So whether or not Fermat did possess a "truly marvelous proof" of his theorem, one that could not fit in the margin of his book, will forever remain his secret. Quote Link to comment Share on other sites More sharing options...

ruveyn1 Posted March 5, 2013 Report Share Posted March 5, 2013 (edited) Given what it took to finally prove FLT is is fairly clear that Fermat did NOT have a proof that would stand up to modern standards of rigor. If you want to see how Wiles did it look at http://www.cs.berkeley.edu/~anindya/fermat.pdf It is only 109 pages long. Fermat claimed his proof fit on the margin of a text book. ruveyn1 Edited March 5, 2013 by ruveyn1 Quote Link to comment Share on other sites More sharing options...

dream_weaver Posted March 6, 2013 Author Report Share Posted March 6, 2013 (edited) Sitting down with a peice of paper after finishing the book, I soon found it filled with numbers. Let me share a few with you. Let's start with the squares. 1^{2}=1, 2^{2}=4, 3^{2}=9, 4^{2}=16, 5^{2}=25, 6^{2}=36, 7^{2}=49, 8^{2}=64, 9^{2}=81, 10^{2}=100, 11^{2}=121, 12^{2}=144, 13^{2}=169, 14^{2}=196, 15^{2}=225, 16^{2}=256, 17^{2}=289, 18^{2}=324, 19^{2}=361, 20^{2}=400 Moving onto the cubes we find, 1^{3}=1, 2^{3}=8, 3^{3}=27, 4^{3}=64, 5^{3}=125, 6^{3}=216, 7^{3}=343, 8^{3}=512, 9^{3}=729, 10^{3}=1000, 11^{3}=1331, 12^{3}=1728, 13^{3}=2197, 14^{3}=2744, 15^{3}=3375, 16^{3}=4096, 17^{3}=4913, 18^{3}=5832, 19^{3}=6859, 20^{3}=8000 Raising to the power of 4 yields, 1^{4}=1, 2^{4}=16, 3^{4}=81, 4^{4}=256, 5^{4}=625, 6^{4}=1296, 7^{4}=2401, 8^{4}=4096, 9^{4}=6561, 10^{4}=10000, 11^{4}=14641, 12^{4}=20736, 13^{4}=28561, 14^{4}=38416, 15^{4}=50625, 16^{4}=65536, 17^{4}=83521, 18^{4}=104976, 19^{4}=130321, 20^{4}=160000 Looking to the power of 5^{ths}, 1^{5}=1, 2^{5}=32, 3^{5}=243, 4^{5}=1024, 5^{5}=3125, 6^{5}=7776, 7^{5}=16807, 8^{5}=32768, 9^{5}=59049, 10^{5}=100000, 11^{5}=161051, 12^{5}=248832, 13^{5}=371293, 14^{5}=537824, 15^{5}=759375, 16^{5}=1048576, 17^{5}=1419857, 18^{5}=1889568, 19^{5}=2476099, 20^{5}=3200000 And lastly at the power of 6^{ths}. 1^{6}=1, 2^{6}=64, 3^{6}=729, 4^{6}=4096, 5^{6}=15625, 6^{6}=46656, 7^{6}=117649, 8^{6}=262144, 9^{6}=531441, 10^{6}=1000000, 11^{6}=1771561, 12^{6}=2985984, 13^{6}=4826809, 14^{6}=7529536, 15^{6}=11390625, 16^{6}=16777216, 17^{6}=24137569, 18^{6}=34012224, 19^{6}=47045881, 20^{6}=64000000 This does give quite a array of numbers. ================================================================================================================================== Let's take this a step further. Starting with the squared differences, 4-1=3, 9-4=5, 16-9=7, 25-16=9, 36-25=11, 49-36=13, 64-49=15, 81-64=17, 100-81=19, 121-100=21, 144-121=23, 169-144=25, 196-169=27, 225-196=29, 256-225=31, 289-256=33, 324-289=35, 361-324=37, 400-361=39 The cubed differences are as follows, 8-1=7, 27-8=19, 64-27=37, 125-64=61, 216-125=91, 343-216=127, 512-343=169, 729-512=217, 1000-729=271, 1331-1000=331, 1728-1331=397, 2197-1728=469, 2744-2197=547, 3375-2744=631, 4096-3375=721, 4913-4096=817, 5832-4913=919, 6859-5832=1027, 8000-6859=1141 The differences raised to the 4^{th} power are, 16-1=15, 81-16=65, 256-81=175, 625-256=369, 1296-625=671, 2401-1296=1105, 4096-2401=1695, 6561-4096=2465, 10000-6561=3439, 14641-10000=4641, 20736-14641=6095, 28561-20736=7825, 38416-28561=9855, 50625-38416=12209, 65536-50625=14911, 83521-65536=17985, 104976-83521=21455, 130321-104976=25345, 160000-130321=29679 The differences raised to the 5^{th} power are, 32-1=31, 243-32=211, 1024-243=781, 3125-1024=2101, 7776-3125=4651, 16807-7776=9031, 32768-16807=15961, 59049-32768=26281, 100000-59049=40951, 161051-100000=61051, 248832-161051=87781, 371293-248832=122461, 537824-371293=166531, 759375-537824=221551, 1048576-759375=289201, 1419857-1048576=371281, 1889568-1419857=469711, 2476099-1889568=586531, 3200000-2476099=723901 And lastly, the differences raised to the 6^{th} power are, 64-1=63, 729-64=665, 4096-729=3367, 15625-4096=11529, 46656-15625=31031, 117649-46656=70993, 262144-117649=144495, 531441-262144=269297, 1000000-531441=468559, 1771561-1000000=771561, 2985984-1771561=1214423, 4826809-2985984=1840825, 7529536-4826809=2702727, 11390625-7529536=3861089, 16777216-11390625=5386591, 24137569-16777216=7360353, 34012224-24137569=9874655, 47045881-34012224=13033657, 64000000-47045881=16954119 ========================================================================================================================================= Summarizing: Squared: 1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23, 25, 27, 29, 31, 33, 35, 37, 39 Cubed: 1, 7, 19, 37, 61, 91, 127, 169, 217, 271, 331, 397, 469, 547, 631, 721, 817, 919, 1027, 1141 Quadratic: 1, 15, 65, 175, 369, 671, 1105, 1695, 2465, 3439, 4641, 6095, 7825, 9855, 12209, 14911, 17985, 21455, 25345, 29679 To the 5^{th}, 1, 31, 211, 781, 2101, 4651, 9031, 15961, 26281, 40951, 61051, 87781, 122461, 166531, 221551, 289201, 371281, 469711, 586531, 723901 To the 6^{th}, 1, 63, 665, 3367, 11529, 31031, 70993, 144495, 269297, 468559, 771561, 1214423, 1840825, 2702727, 3861089, 5386591, 7360353, 9874655, 13033657, 16954119 ========================================================================================================================================= Reviewing the squared sequence, the assigned difference can be represented as:2*X+1. 2*0+1=1, 2*1+1=3 2*2+1=5, 2*3+1=7, 2*4+`=9, 2=5+1=11, 2*6+1=13, 2*7+1=15, 2*8+1=17, 2*9+1=19, 2*10+1=21, etc. What is not as obvious is for the cubed sequence: X^{2}+(X+1)(2*X+1) 0^{2}+(0+1)(2*0+1)=1, 1^{2}+(1+1)(2*1+1)=7, 2^{2}+(1+1)(2*1+1)=7, 2^{2}+(2+1)(2*2+1)=19, 3^{2}+(3+1)(2*3+1)=37, 4^{2}+(4+1)(2*4+1)=61, etc. Stepping to the next sequence, what becomes clearer is X^{3}+((X+1)(X^{2}+(X+1)(2*X+1)). This gives us: 0^{3}+(0+1)(0^{2}+(0+1)(2*0+1))=1, 1^{3}+(1+1)(1^{2}+(1+1)(2*1+1))=15, 2^{3}+(2+1)(2^{2}+(2+1)(2*2+1))=65, 3^{3}+(3+1)(3^{2}+(3+1)(2*3+1))=175, etc. Extrapolating to the 5^{th} power: X^{4}+(X+1)((X^{3}+(X+1)((X^{2}+(X+1)(2*X+1))) yields: 0^{4}+(0+1)(0^{3}+(0+1)((0^{2}+(0+1)(2*0+1)))=1, 1^{4}+(1+1)(1^{3}+(1+1)((1^{2}+(x1+1)(2*1+1))=31, 2^{4}+(2+1)(2^{3}+(2+1)((2^{2}+(2+1)(2*2+1)))=211, 3^{4}+(3+1)(3^{3}+(3+1)((3^{2}+(3+1)(2*3+1)))=671, etc. And for this example, taking it finally to the 6^{th} power: X^{5}+(X+1)((X^{4}+(X+1)((X^{3}+(X+1)((X^{2}+(X+1)(2*X+1)))) 0^{5}+(0+1)(0^{4}+(0+1)(0^{3}+(0+1)(0^{2}+(0+1)(2*0+1))))=1, 1^{5}+(1+1)(1^{4}+(1+1)(1^{3}+(1+1)(1^{2}+(1+1)(2*1+1))))=63, 2^{5}+(2+1)(2^{4}+(2+1)(2^{3}+(2+1)(X^{2}+(2+1)(2*2+1))))=665, 3^{5}+(3+1)(3^{4}+(3+1)(3^{3}+(3+1)(3^{2}+(3+1)(2*3+1))))=3367, etc. Could Fermat have induced from a pattern such as this, that for a power greater than 2, A^{n}+B^{n}=C^{n} could not have been resolved into an integral relationship? fixed 761 from 671 on the 5th power 3^{4,}, & parenthesis fixed on various others. Edited March 6, 2013 by dream_weaver Quote Link to comment Share on other sites More sharing options...

ruveyn1 Posted March 6, 2013 Report Share Posted March 6, 2013 Sitting down with a peice of paper after finishing the book, I soon found it filled with numbers. Let me share a few with you. Let's start with the squares. 1^{2}=1, 2^{2}=4, 3^{2}=9, 4^{2}=16, 5^{2}=25, 6^{2}=36, 7^{2}=49, 8^{2}=64, 9^{2}=81, 10^{2}=100, 11^{2}=121, 12^{2}=144, 13^{2}=169, 14^{2}=196, 15^{2}=225, 16^{2}=256, 17^{2}=289, 18^{2}=324, 19^{2}=361, 20^{2}=400 Moving onto the cubes we find, 1^{3}=1, 2^{3}=8, 3^{3}=27, 4^{3}=64, 5^{3}=125, 6^{3}=216, 7^{3}=343, 8^{3}=512, 9^{3}=729, 10^{3}=1000, 11^{3}=1331, 12^{3}=1728, 13^{3}=2197, 14^{3}=2744, 15^{3}=3375, 16^{3}=4096, 17^{3}=4913, 18^{3}=5832, 19^{3}=6859, 20^{3}=8000 Raising to the power of 4 yields, 1^{4}=1, 2^{4}=16, 3^{4}=81, 4^{4}=256, 5^{4}=625, 6^{4}=1296, 7^{4}=2401, 8^{4}=4096, 9^{4}=6561, 10^{4}=10000, 11^{4}=14641, 12^{4}=20736, 13^{4}=28561, 14^{4}=38416, 15^{4}=50625, 16^{4}=65536, 17^{4}=83521, 18^{4}=104976, 19^{4}=130321, 20^{4}=160000 Looking to the power of 5^{ths}, 1^{5}=1, 2^{5}=32, 3^{5}=243, 4^{5}=1024, 5^{5}=3125, 6^{5}=7776, 7^{5}=16807, 8^{5}=32768, 9^{5}=59049, 10^{5}=100000, 11^{5}=161051, 12^{5}=248832, 13^{5}=371293, 14^{5}=537824, 15^{5}=759375, 16^{5}=1048576, 17^{5}=1419857, 18^{5}=1889568, 19^{5}=2476099, 20^{5}=3200000 And lastly at the power of 6^{ths}. 1^{6}=1, 2^{6}=64, 3^{6}=729, 4^{6}=4096, 5^{6}=15625, 6^{6}=46656, 7^{6}=117649, 8^{6}=262144, 9^{6}=531441, 10^{6}=1000000, 11^{6}=1771561, 12^{6}=2985984, 13^{6}=4826809, 14^{6}=7529536, 15^{6}=11390625, 16^{6}=16777216, 17^{6}=24137569, 18^{6}=34012224, 19^{6}=47045881, 20^{6}=64000000 This does give quite a array of numbers. ================================================================================================================================== Let's take this a step further. Starting with the squared differences, 4-1=3, 9-4=5, 16-9=7, 25-16=9, 36-25=11, 49-36=13, 64-49=15, 81-64=17, 100-81=19, 121-100=21, 144-121=23, 169-144=25, 196-169=27, 225-196=29, 256-225=31, 289-256=33, 324-289=35, 361-324=37, 400-361=39 The cubed differences are as follows, 8-1=7, 27-8=19, 64-27=37, 125-64=61, 216-125=91, 343-216=127, 512-343=169, 729-512=217, 1000-729=271, 1331-1000=331, 1728-1331=397, 2197-1728=469, 2744-2197=547, 3375-2744=631, 4096-3375=721, 4913-4096=817, 5832-4913=919, 6859-5832=1027, 8000-6859=1141 The differences raised to the 4^{th} power are, 16-1=15, 81-16=65, 256-81=175, 625-256=369, 1296-625=671, 2401-1296=1105, 4096-2401=1695, 6561-4096=2465, 10000-6561=3439, 14641-10000=4641, 20736-14641=6095, 28561-20736=7825, 38416-28561=9855, 50625-38416=12209, 65536-50625=14911, 83521-65536=17985, 104976-83521=21455, 130321-104976=25345, 160000-130321=29679 The differences raised to the 5^{th} power are, 32-1=31, 243-32=211, 1024-243=781, 3125-1024=2101, 7776-3125=4651, 16807-7776=9031, 32768-16807=15961, 59049-32768=26281, 100000-59049=40951, 161051-100000=61051, 248832-161051=87781, 371293-248832=122461, 537824-371293=166531, 759375-537824=221551, 1048576-759375=289201, 1419857-1048576=371281, 1889568-1419857=469711, 2476099-1889568=586531, 3200000-2476099=723901 And lastly, the differences raised to the 6^{th} power are, 64-1=63, 729-64=665, 4096-729=3367, 15625-4096=11529, 46656-15625=31031, 117649-46656=70993, 262144-117649=144495, 531441-262144=269297, 1000000-531441=468559, 1771561-1000000=771561, 2985984-1771561=1214423, 4826809-2985984=1840825, 7529536-4826809=2702727, 11390625-7529536=3861089, 16777216-11390625=5386591, 24137569-16777216=7360353, 34012224-24137569=9874655, 47045881-34012224=13033657, 64000000-47045881=16954119 ========================================================================================================================================= Summarizing: Squared: 1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23, 25, 27, 29, 31, 33, 35, 37, 39 Cubed: 1, 7, 19, 37, 61, 91, 127, 169, 217, 271, 331, 397, 469, 547, 631, 721, 817, 919, 1027, 1141 Quadratic: 1, 15, 65, 175, 369, 671, 1105, 1695, 2465, 3439, 4641, 6095, 7825, 9855, 12209, 14911, 17985, 21455, 25345, 29679 To the 5^{th}, 1, 31, 211, 781, 2101, 4651, 9031, 15961, 26281, 40951, 61051, 87781, 122461, 166531, 221551, 289201, 371281, 469711, 586531, 723901 To the 6^{th}, 1, 63, 665, 3367, 11529, 31031, 70993, 144495, 269297, 468559, 771561, 1214423, 1840825, 2702727, 3861089, 5386591, 7360353, 9874655, 13033657, 16954119 ========================================================================================================================================= Reviewing the squared sequence, the assigned difference can be represented as:2*X+1. 2*0+1=1, 2*1+1=3 2*2+1=5, 2*3+1=7, 2*4+`=9, 2=5+1=11, 2*6+1=13, 2*7+1=15, 2*8+1=17, 2*9+1=19, 2*10+1=21, etc. What is not as obvious is for the cubed sequence: X^{2}+(X+1)(2*X+1) 0^{2}+(0+1)(2*0+1)=1, 1^{2}+(1+1)(2*1+1)=7, 2^{2}+(1+1)(2*1+1)=7, 2^{2}+(2+1)(2*2+1)=19, 3^{2}+(3+1)(2*3+1)=37, 4^{2}+(4+1)(2*4+1)=61, etc. Stepping to the next sequence, what becomes clearer is X^{3}+((X+1)(X^{2}+(X+1)(2*X+1)). This gives us: 0^{3}+(0+1)(0^{2}+(0+1)(2*0+1))=1, 1^{3}+(1+1)(1^{2}+(1+1)(2*1+1))=15, 2^{3}+(2+1)(2^{2}+(2+1)(2*2+1))=65, 3^{3}+(3+1)(3^{2}+(3+1)(2*3+1))=175, etc. Extrapolating to the 5^{th} power: X^{4}+(X+1)((X^{3}+(X+1)((X^{2}+(X+1)(2*X+1))) yields: 0^{4}+(0+1)(0^{3}+(0+1)((0^{2}+(0+1)(2*0+1)))=1, 1^{4}+(1+1)(1^{3}+(1+1)((1^{2}+(x1+1)(2*1+1))=31, 2^{4}+(2+1)(2^{3}+(2+1)((2^{2}+(2+1)(2*2+1)))=211, 3^{4}+(3+1)(3^{3}+(3+1)((3^{2}+(3+1)(2*3+1)))=671, etc. And for this example, taking it finally to the 6^{th} power: X^{5}+(X+1)((X^{4}+(X+1)((X^{3}+(X+1)((X^{2}+(X+1)(2*X+1)))) 0^{5}+(0+1)(0^{4}+(0+1)(0^{3}+(0+1)(0^{2}+(0+1)(2*0+1))))=1, 1^{5}+(1+1)(1^{4}+(1+1)(1^{3}+(1+1)(1^{2}+(1+1)(2*1+1))))=63, 2^{5}+(2+1)(2^{4}+(2+1)(2^{3}+(2+1)(X^{2}+(2+1)(2*2+1))))=665, 3^{5}+(3+1)(3^{4}+(3+1)(3^{3}+(3+1)(3^{2}+(3+1)(2*3+1))))=3367, etc. Could Fermat have induced from a pattern such as this, that for a power greater than 2, A^{n}+B^{n}=C^{n} could not have been resolved into an integral relationship? fixed 761 from 671 on the 5th power 3^{4,}, & parenthesis fixed on various others. It does not follow. Just because something fails for the first gazillion does not mean it will not succeed for gazillion plus 1. You cannot, in general disprove non existence of something just by exhibiting a finite number of failures. Mathematicians very rarely work by empirical induction (not to be confused with arithmetic induction which is something completely different). Quote Link to comment Share on other sites More sharing options...

dream_weaver Posted March 6, 2013 Author Report Share Posted March 6, 2013 Using 2x+1 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20 <--- x 1, 4, 9, 16, 25, 36, 49, 64, 81, 100, 121, 144, 169, 196, 225, 256, 289, 324, 361, 400 <--- x^{2} 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23, 25, 27, 29, 31, 33, 35, 37, 39 <--- 2x+1 Little was stated in the book about finding Pythagorian triples. Wikipedia and Wolfram have some formulas that can be used. I've not tried to contrast what I've done here with what has been done there at this point. From the last row, the 9 appears between the 16 and 25 of the second row. 3,4,5 Next we find 25 between 144 and 169 yielding another triple of 5,12,13 To find what works with 7, 2*x+1=49, x would be between 24 and 25 yielding 7,24,25 Starting with 3,4,5 we also can see that 2(3,4,5) = 6,8,10 and 3(3,4,5) = 9,12,15 and 4(3,4,5) = 12,16,20 This can also be done with our earlier triplet using 2(5,12,13) to get 10,24,26 While certainly not exhaustive, there are some definite patterns available to be explored. Moving on to x^{2}+(x+1)(2x+1) we get the following: 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20 <--- x 1, 8, 27, 64, 125, 216, 343, 512, 729, 1000, 1331, 1728, 2197, 2744, 3375, 4096, 4913, 5832, 6859, 8000 <--- x^{3} 7, 19, 37, 61, 91, 127, 169, 217, 271, 331, 397, 469, 547, 631, 721, 817, 919, 1027, 1141 <--- x^{2}+(x+1)(2x+1) x^{3}+(x+1)(x^{2}+(x+1)(2x+1)) 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20 <--- x 1, 16, 81, 256, 625, 1296, 2401, 4096, 6561, 10000, 14641, 20736, 28561, 38416, 50625, 65536, 83521, 104976, 130321, 160000 <---x^{4} 15, 65, 175, 369, 671, 1105, 1695, 2465, 3439, 4641, 6095, 7825, 9855, 12209, 14911, 17985, 21455, 25345, 29679 <--- x^{3}+(x+1)(x^{2}+(x+1)(2x+1)) Perhaps noting the number of prime numbers in the cubed series this may have inspired Fermat to develop his formula for primes, 2^{2^n}+1, which was discovered to be false. The fact is, a pattern can be isolated in the differences, and it can be formulated along the lines of x^{n}+(x+1)( . . . .(x^{2}+(x+1)(2x+1))). Arranging the results in a way that other patterns emerged for the squared series failed to provide similar patterns for myself for up to x^{8} for the first 20 integers before excel started converting the answers into scientific notation. What Fermat described as a "truly marvelous proof" that would not fit in the margin where he made his notation, has not been discovered. Quote Link to comment Share on other sites More sharing options...

ruveyn1 Posted March 6, 2013 Report Share Posted March 6, 2013 What Fermat described as a "truly marvelous proof" that would not fit in the margin where he made his notation, has not been discovered. It probably does not exist. Quote Link to comment Share on other sites More sharing options...

John Link Posted March 7, 2013 Report Share Posted March 7, 2013 Given what it took to finally prove FLT is is fairly clear that Fermat did NOT have a proof that would stand up to modern standards of rigor. If you want to see how Wiles did it look at http://www.cs.berkeley.edu/~anindya/fermat.pdf It is only 109 pages long. Fermat claimed his proof fit on the margin of a text book. ruveyn1 I have not read the proof cited above, but I consider it possible that Fermat might have had, or someone else might discover, a much more elegant and insightful proof that is much shorter than 109 pages. I'm sure it wouldn't be the first time that a long proof could be replaced by a much shorter one, but I don't have any examples at hand to cite. Quote Link to comment Share on other sites More sharing options...

ruveyn1 Posted March 7, 2013 Report Share Posted March 7, 2013 I have not read the proof cited above, but I consider it possible that Fermat might have had, or someone else might discover, a much more elegant and insightful proof that is much shorter than 109 pages. I'm sure it wouldn't be the first time that a long proof could be replaced by a much shorter one, but I don't have any examples at hand to cite. I really doubt it. Some of the greatest mathematicians of all time have had a go at it, and Wiles was the one who finally did it after 300+ years. I will believe there is a short elementary proof when I see it with my own eyes.l Quote Link to comment Share on other sites More sharing options...

dream_weaver Posted March 7, 2013 Author Report Share Posted March 7, 2013 Considering that Aristotle may have induced his rules of logic methodology from pre-Euclidean geometry, the application of todays modern standards of rigor within the science of mathematics sometimes elude to questions I have difficulty discovering how to articulate. Cubes, Quads, Quints, etc, introduce methods of using number to measure number differently than than the number line. articulates. Being enumerated among the greatest mathematicians of all times does not ensure adherence to objectivity. While Wiles finally offered a proof that is acceptable to the mathematical community, does not ensure that it is indeed objective, in an Objective sense of the term. Quote Link to comment Share on other sites More sharing options...

dream_weaver Posted March 8, 2013 Author Report Share Posted March 8, 2013 (edited) Fermat’s Last Theorem Pythagrean Theorem states that the hypotenuse of a right triangle can be resolved by taking the square root of the sum of the sides. Pythagrean triples have been discovered and used historically. Fermat claimed in a margin of a lost book that there is no integer solution for A^{n>2+}B^{n>2}=C^{n>2} I now agree with him, and not for the fact that a 109 page dissertation was written on the subject that I have not read. The diagonal of a square involves the square root of 2. If a side is an integer, the diagonal is a factor involving the square root of 2. If the diagonal is an integer, the sides will have a factor involving the square root of 2. To solve for the greater diagonal on a cube also invokes the square root of 2. The formula breaks down to square root of 3*x^{2}. This can be reduced to the square root of 3, also an irrational number. Consider a cube with a side of 1, the diagonal is the square root of 3. Consider a cube with a side of 2, the diagonal is 2*square root of 3. The formula works out to (x^{2}+x^{2}+x^{2})^{1/2} To end up with an integer for the answer, the side would have to be an irrational number. The formula for exponents 3 or greater. Square root of (x^{2}+x^{2}+x^{(x-2)}) (For powers greater than 3, the result ends up with multiple cubes. 2^{4 }would be 2 cubes of 8 for a total of 16. 8^{8} would be 32768 cubes of 4096) It doesn’t matter whether there is 1 cube, or a gazillion cubes, the answer remains that a diagonal of a cube contains a factor of the square root of 3, or an integer was not used to derive the answer. Edited March 8, 2013 by dream_weaver Quote Link to comment Share on other sites More sharing options...

ruveyn1 Posted March 8, 2013 Report Share Posted March 8, 2013 It doesn’t matter whether there is 1 cube, or a gazillion cubes, the answer remains that a diagonal of a cube contains a factor of the square root of 3, or an integer was not used to derive the answer. The real task was showing that x^n + y^n = z^n did not have nonzero solutions x, y, z for ALL n > 2. That stymied the world's best mathematicians for over 300 years. Here is a problem which if you can solve it will make you as famous as Dr. Wiles. It is the proof of the Collatz conjecture. I will define a function on the integers as follows: T(n) = n/2 if n is even T(n) = (3*n + 1)/2 if n is odd. Here is the problem: show that starting with any integer N if you keep applying the function T you will eventually get to 1. Example.. N = 7 7, 11, 17, 26, 13, 20, 10, 5, 4, 2, 1 That problem has been open since around 1950. Paul Erdos one of the great great mathematicians was of the opinion that mankind was not ready to solve that problem. It has already been show for N < 10^20 this is true, but a proof for ALL N remains to be found. All attempts to prove that the general proposition is undecidable have also failed. ruveyn1 Quote Link to comment Share on other sites More sharing options...

dream_weaver Posted March 8, 2013 Author Report Share Posted March 8, 2013 (edited) (x^{2}+x^{2}+x^{x-2})^{1/2 }should show that x^{n}+y^{n}=z^{n} has no nonzero solutions x,y,z for ALL n>2. I do not have the mathematical forte to convert it at this time. It is true, for the same reason that the hypotenuse of a square is incommensurate to its sides. I have a bookmark for consideration of Collatz conjecture. If fame were my goal, . . . who knows. By the way, 5-->(3*n+1/2)=8, not 4, although 8, 4, 2, 1 does finish the hypothesis. Edited March 8, 2013 by dream_weaver Quote Link to comment Share on other sites More sharing options...

ruveyn1 Posted March 8, 2013 Report Share Posted March 8, 2013 (edited) (x^{2}+x^{2}+x^{x-2})^{1/2 }should show that x^{n}+y^{n}=z^{n} has no nonzero solutions x,y,z for ALL n>2. I do not have the mathematical forte to convert it at this time. It is true, for the same reason that the hypotenuse of a square is incommensurate to its sides. I have a bookmark for consideration of Collatz conjecture. If fame were my goal, . . . who knows. By the way, 5-->(3*n+1/2)=8, not 4, altsohough 8, 4, 2, 1 does finish the hypothesis. sorry about the error. A trivial corollary: once a collatz sequence hits a power of two, the game is over. You say you have a proof : "(x^{2}+x^{2}+x^{x-2})^{1/2 }should show that x^{n}+y^{n}=z^{n} has no nonzero solutions x,y,z for ALL n>2. I do not have the mathematical forte to convert it at this time. It is true, for the same reason that the hypotenuse of a square is incommensurate to its sides." Claims are cheap. Valid proofs are dear: Show the proof or be still. Edited March 8, 2013 by ruveyn1 Quote Link to comment Share on other sites More sharing options...

dream_weaver Posted March 8, 2013 Author Report Share Posted March 8, 2013 Do the math. Quote Link to comment Share on other sites More sharing options...

dream_weaver Posted March 8, 2013 Author Report Share Posted March 8, 2013 (edited) Pehaps this picture will help. I'll have to upload it later. Edited March 8, 2013 by dream_weaver Quote Link to comment Share on other sites More sharing options...

ruveyn1 Posted March 8, 2013 Report Share Posted March 8, 2013 Do the math. No. YOU do the math. You claimed the proof, now show it step by step so we can check to see if it is right. Quote Link to comment Share on other sites More sharing options...

dream_weaver Posted March 8, 2013 Author Report Share Posted March 8, 2013 (edited) Alright. Now lets see if we can attach this thing. (Between here) (here)(here) Its hard to see (and here.) Imagine walking into a building at the door. The building is X by X. Walk across the diagonal to the corner opposite the door. Climb the stairwell crossing back across the diagonal to the corner over the door. This is (3X^{2})^{^1/2} The next solution where I corrected my original formula, is for a 2 story, 3 story, 4 story etc. The stairwell starts in the same corner, only it climbs to the top corner over the door of the destination floor . (X^{2}+X^{2}+((exponent-2)X)^{2})^{1/2} Edited March 8, 2013 by dream_weaver Quote Link to comment Share on other sites More sharing options...

dream_weaver Posted March 8, 2013 Author Report Share Posted March 8, 2013 (edited) As to Collantz Conjecture that should work for any odd number. T(n) = n/2 if n is even T(n) = (3*n + 1)/2 if n is odd. T(n) = n/2 if n is even T(n) = (5*n + 1)/2 if n is odd. T(n) = n/2 if n is even T(n) = (7*n + 1)/2 if n is odd. T(n) = n/2 if n is even T(n) = (9*n + 1)/2 if n is odd. ... Edited March 8, 2013 by dream_weaver Quote Link to comment Share on other sites More sharing options...

ruveyn1 Posted March 8, 2013 Report Share Posted March 8, 2013 As to Collantz Conjecture that should work for any odd number. T(n) = n/2 if n is even T(n) = (3*n + 1)/2 if n is odd. T(n) = (5*n + 1)/2 if n is odd. T(n) = (7*n + 1)/2 if n is odd. T(n) = (9*n + 1)/2 if n is odd. ... Prove that and you will become an Immortal. Quote Link to comment Share on other sites More sharing options...

ruveyn1 Posted March 8, 2013 Report Share Posted March 8, 2013 Alright. Now lets see if we can attach this thing. FermatsLastTheorem.jpg Imagine walking into a building at the door. The building is X by X. Walk across the diagonal to the corner opposite the door. Climb the stairwell crossing back across the diagonal to the corner over the door. This is (3X^{2})^{^1/2} The next solution where I corrected my original formula, is for a 2 story, 3 story, 4 story etc. The stairwell starts in the same corner, only it climbs to the top corner over the door of the destination floor . (X^{2}+X^{2}+((exponent-2)X)^{2})^{1/2} Do you have any idea of what constitutes a rigorous proof? Quote Link to comment Share on other sites More sharing options...

dream_weaver Posted March 8, 2013 Author Report Share Posted March 8, 2013 (edited) Nope. I had a little pre-calc 30 years ago. But sooner or later, (3n+1)=2^{x}, as does (5n+1) or (7n+1) 3n+1 does for 1, 5 and more, 5n+1 does for 3, 7 and more. 7n+1 does for 1, 5 and more. I would tend to believe that is why it works. Edited March 8, 2013 by dream_weaver Quote Link to comment Share on other sites More sharing options...

ruveyn1 Posted March 8, 2013 Report Share Posted March 8, 2013 Nope. I had a little pre-calc 30 years ago. Do two things. 1 Learn what a proof is 2. Write you proof out carefully and have a professional mathematician vet your proof for errors. If you can prove FLT by elementary means and you are under 45 years of age, I guarantee you will get a Fields Medal. Quote Link to comment Share on other sites More sharing options...

dream_weaver Posted March 8, 2013 Author Report Share Posted March 8, 2013 Already past that point. I just enjoy tinkering with this stuff occasionally. The drawing I produced is more in line with my forte. I have to envision what it looks like in order to create it. Quote Link to comment Share on other sites More sharing options...

dream_weaver Posted March 9, 2013 Author Report Share Posted March 9, 2013 (edited) The more I look at my final formula using the exponent, I do not think it accurate represents the number of floors. I think there are many more, and are influenced by the number being exponented in some way.. Up to the (X^{2}+X^{2}+X^{2}) is the means of deriving that figure. Like the pythagorean triple cannot occur in a square, the cube exhibits similar properties concretized in this form. Edited March 9, 2013 by dream_weaver Quote Link to comment Share on other sites More sharing options...

dream_weaver Posted March 9, 2013 Author Report Share Posted March 9, 2013 (edited) Another way of looking at the problem. Take a cube of 10 X 10 X 10 boxes, Take 8 X 8 X 8 boxes away, Take 2 X 2 X 2 boxes away. What is left? 10^{3}-8^{3}-2^{3 }cannot be 0 The remainder is not enough to make 8^{3} and 3^{3 }or 9^{3} and 2^{3}. (A+B )^{3}-A^{3}-B^{3}>0 (A+1)^{3}+B^{3}>(A+B )^{3} A^{3}+(B+1)^{3}>(A+B )^{3} Edited March 9, 2013 by dream_weaver Quote Link to comment Share on other sites More sharing options...

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